Math 500: Topology

Homework 8
Lawrence Tyler Rush
<me@tylerlogic.com>
January 12, 2013
http://coursework.tylerlogic.com/courses/math500/homework08

Problems

P-1 Munkres §51 exercise 3 b, c, and d

(b) Contractible spaces are path connected.

Assume X to be a contractible space. Let H be the homotopy between the identity function on X and the constant function c. Letting x,y be elements of X, define the map fxy : [0,1] → X by
{ 1 fxy(t) = H (x,2t) t ∈ [01,∕2] H (y,2(1- t)) t ∈ [ ∕2,1]

This map is well-defined since H(x,2(1 ∕ 2)) = H(x,1) = c and H(y,2(1 -1 ∕ 2)) = H(y,1) = c according to homotopic nature of H. Furthermore, due to H’s continuity fxy is continuous by the pasting lemma. Thus since the domain, [0,1], of fxy is a closed interval and both fxy(0) = H(x,2(0)) = H(x,0) = idX(x) = x and fxy(1) = H(y,2(1 - 1)) = H(y,0) = idX(y) = y then f is a path between x and y. So X is path connected.

(c) [X,Y ] contains one element for all X when Y is contractible

Assume that Y is contractible. Showing that [X,Y ] has only one element amounts to showing that any two continuous maps from X to Y are homotopic. We proceed thusly.

Let f and g be continuous functions from X to Y . We create a homotopy between them by using the path connectedness of Y , by way of the previous problem. Define H′ : X × [0,1] → Y by

{ 1 H ′(x,t) = H(f(x),2t) t ∈ [01,∕2] H(g(x),2(1- t)) t ∈ [ ∕2,1]

where H is the homotopy between the identity map on Y and some constant function c on Y . As previously, this function is well defined at t = 1 ∕ 2 since H′(f(x),2(1 ∕ 2)) = H(f(x),1) = c and H′(g(x),2(1 -1 ∕ 2)) = H(g(x),1) = c. Also, H′ is continuous by the pasting lemma. Thus because H′(x,0) = H(f(x),0) = idY (f(x)) = f(x) and H′(x,1) = H(g(x),0) = idY (g(x)) = g(x) then H′ is a homotopy between f and g. Thus f ≃ g, implying that all continuous elements of Y X are homotopic. So [X,Y ] has a size of one.

(d) Show [X,Y ] has one element for contractible X and path connected Y

Assume that X is a contractible space and Y is path connected. Let f,g be continuous maps from X to Y and H be the homotopy between idX and some constant map c on X. Since Y is path connected we can find a path h between f(c) and g(c). We can manipulate the domain of h as we see fit, so let’s make it [1 ∕ 3,2 ∕ 3]. With these things, we construct a homotopy essentially by contracting f(x) to f(c) with H, using h to travel to g(c), and then expanding g(c) to g(x) again using H. We do this by defining H′ as
({ f(H (x,3t)) t ∈ [0,1∕3] H ′(x,t) = h(t) t ∈ [1∕3,2∕3] ( g(H (x,3(1- t))) t ∈ [2∕3,1]

This map is well defined since f(H(x,3(1 ∕ 3))) = f(H(x,1)) = f(c) = h(1 ∕ 3) and g(H(x,3(1 -2 ∕ 3))) = g(H(x,1)) = g(c) = h(2 ∕ 3). Furthermore H′ is continuous by the pasting lemma since its constituent parts are either continuous or compositions of continuous functions. Thus f ≃ g via the homotopy of H′. Since f and g were abitrary, then all continous functions from X to Y are homotopic to each other, and therefore [X,Y ] has a single element.